Lecture 14 Chapter 19 Ideal Gas Law and Kinetic Theory of Gases. Chapter 20 Entropy and the Second Law of Thermodynamics

Transcription

1 Lecture 14 Chapter 19 Ideal Gas Law and Kinetic Theory of Gases Now we to look at temperature, pressure, and internal energy in terms of the motion of molecules and atoms? Relate to the 1st Law of Thermodynamics Chapter 20 Entropy and the Second Law of Thermodynamics

2 Thermal expansion cracking the nut. (Strength of electrical forces) Jug O' Air (Inflate with bike pump and watch temp. rise) p/t = constant Boiling by Cooling (Ice on beaker) Boiling by Reducing Pressure(Vacuum in Bell jar) Dipping Duck Toy wet head cools as water evaporates off it - pressure drops inside head contained pressure pushes water up tube when center of gravity is exceeded head tips exposes bottom of tube then pressure equalized Leslie cube and the laser themometer.

3 Pressure reduction due to cooling inside coke can crushing it when placed in water. pv=nrt, p/t=constant

4 Thermal effects using liquid nitrogen again. For air in the balloon at room temp At room temp we have pv=nrt and T=273K then dip it liquid nitrogen p V =nrt with T= 78K, p V should be smaller by a factor of 273/78=3.5 compared to pv So why does the balloon get a lot smaller? Since air boils at 90 K, air is liquid at 78 K and is no longer a gas so the ideal gas law does not apply.)

5 Avogadro s Number How many molecules are there in a cubic meter of air at STP? First, find out how much mass there is. The mass is the density times the volume. m = 1.2 kg/m 3 x 1 m 3 = 1.2 kg. (or 2.5 lbs) Now find the number of moles in the cubic meter and multiply by Avogadro s number 1 m

6 Avogadro s Number N A =6.02 x atoms or molecules in one mole of any gas. One mole is the molecular weight in grams. It is also called the molar mass. 1 mole of air contains 29 gms Then the number of moles in 1.2 kg is n = 1200gm 1 29gm /1mol = 41.3mol Then the number of molecules or atoms is N = 41.3N A = Also 1 / N A = gms Mass of the proton= gms Advogado s number is also related to the Ideal Gas law.

7 Ideal Gases Experiment shows that 1 mole of any gas, such as helium, air, hydrogen, etc at the same volume and temperature has almost the same pressure. At low densities the pressures become even closer and obey the Ideal Gas Law: p = nrt / V V = volume in units of m 3 n = number of moles T = temperature in units of K R = 8.31J/moles K p = pressure(absolute) in units of Pa(N/m 2 ) Called the gas constant

8 Ideal Gas Law in terms of Boltzman Constant p = nrt / V R = N A k nr = nn A k = Nk k = J / K = Boltzman's constant pv = NkT At high gas densities you must add the Van der Waal corrections where a and b are constants. p = NkT V b ( a V )2 Lets use the ideal gas law to understand the 1st law of Thermodynamics ΔE int = Q W

9 Work done by an ideal gas Always keep this picture in your head. Memorize it!! W = W = W = Fdx F Adx A pdv ΔE int = Q W

10 pv diagram For an isothermal process or constant temperature. This is an equation of a hyperbola. pv = nrt p = nrt V = constant V

11 What is the work done by an ideal gas when the temperature is constant? pv = nrt For constant temperature. W = W = V f V i V f V i constant pdv nrt V dv W = nrt V f V i dv V V = nrt [ lnv ] f Vi W = nrt ln V f V i

12 What is the work done by an ideal gas when the temperature is constant? pv = nrt W = nrt ln V f V i Isothermal expansion V f > V i W > 0 ln is positive Isothermal compression V f < V i W < 0 ln is negative Constant volume V f = V i W = 0 ln 1=0 ΔE int = Q W

13 Four situations where the work done by an ideal gas is very clear. W = V f V i pdv For a constant volume process: W = 0 For a constant pressure process W = p(v f V i) For a constant temperature process W = nrt ln V f V i For an ideal gas undergoing any reversible thermodynamic process. W = nr V f V i T V dv

14 Sample problem 19-2 One mole of oxygen expands from 12 to 19 liters at constant temperature of 310 K. What is the work done? W = nrt ln V f n = 1mol V i R = 8.31J / mol K T = 310K V i = 12l V f = 19l (Note volume units cancel) W = (1)(8.31)(310)ln( ) = 1180J W = The area under the curve. ΔE int = Q W

15 What s Next W = nr V f V i T V dv ΔE int = Q W We know how to find W. If we can find the change in internal energy, then we know Q. Resort to a microscopic or kinetic theory of a gas.

16 Kinetic theory model of a gas: Find p due to one molecule first and then sum them up What is the connection between pressure and speed of molecules? We want to find the x component of force per unit area F x = ΔP x Δt ΔP = P f P i ΔP x = ( mv x ) (mv x ) = 2mv x Δt = 2L / v x x component of momentum given to wall is - ΔP x F x = ΔP x Δt = 2mv x = mv 2 x 2L / v x L p = F x L = mv 2 x / L 2 L 2 = m(v 2 x ) / L 3 due to one molecule

17 Find pressure due to all molecules p = F x L = mv 2 x / L 2 L 2 p = m(v 2 2 1x + v 2 x L 3 = m(v 2 x ) / L ) = mnn (v 2 A x ) avg L 3 Due to one molecule Due to all molecules p = nm (v 2 x ) avg V where M is the molar mass and V is L 3 Now find the average molecular velocity.

18 What is the root mean square of the velocity of the Molecules? p = nm (v 2 x ) avg V 2 v avg = (v x 2 ) avg + (v y 2 ) avg + (v z 2 ) avg and (v x 2 ) avg = (v y 2 ) avg = (v z 2 ) avg 2 v avg = 3(v 2 x ) avg (v 2 x ) avg = v 2 avg then 3 p = nm (v2 ) avg 3V Define the root mean square of v v rms = (v 2 ) avg pv = nrt p = nmv 2 rms 3V 2 v rms = 3pV nm = 3nRT nm = 3RT M v rms = 3RT M

19 Average Molecular speeds at 300 K v rms = 3RT M R = 8.31J / mol T = 300K M = 0.002kg Gas m/s Hydrogen 1920 Helium 1370 Water vapor 645 Nitrogen 517 Oxygen 483 v rms = 3(8.31)(300) v rms = 3(8.31)(300) =1933 = 483 Notice that the speed decreases with mass What is the kinetic energy of the molecules?

20 Average Kinetic Energy of the Molecule K avg = 1 mv 2 2 avg = 1 mv 2 2 rms v rms = 3RT M K avg = 1 2 m 3RT M = 3RT 2N A K avg = 3 2 kt K avg = (1.5)( J / K) (293K) = J at room temperature. All ideal gas molecules have the same translational energy at a given T independent of their mass. Remarkable result.

21 Mean Free Path mean free path = average distance molecules travel in between collision λ = 1 2πd 2 N /V d is the diameter of the molecule N/V is the density of molecules

22 Problem. Suppose we have a oxygen molecule at 300 K at p =1 atm with a molecular diameter of d= 290 pm. What is λ,v, and f where f is defined as the frequency of collisions in an ideal gas? f = speed of the molecule/mean free path = v /λ Find λ 1 λ = 2πd 2 N /V pv = NkT N V = p kt λ = kt 2πd 2 p = 1.38x10 23 J / K(300K) 1.414(3.14)(290x10 12 m) 2 (1.01x10 5 Pa) λ = m = 380d

23 What is v, the speed of a molecule in an ideal gas? Use the rms speed. v rms = 3RT M v rms = 3(8.31)(300) = 483m / s What is f, the frequency of collision? f = f = v rms / λ = / s What is the time between collisions? t = 1 f = / s = = 2.2ns

24 Maxwell s speed distribution law: Explains boiling. M P(v) = 4π( 2πRT ) 3 0 P(v)dv =1 2 v 2 e Mv2 RT v p Area under red or green curve = 1 Note there are three velocities v avg = 0 vp(v) v rms = (v 2 ) avg = v 2 P(v)dv v p is the most probable speed It is the faster moving molecules in the tail of the distribution that escape from the surface of a water. 0 v p

25 Want to relate ΔE int to the kinetic energy of the atoms of the gas. Assume we have a monoatomic gas. Only have translational energy. No rotational energy. No vibrational energy. Neglect binding energy of electrons. No changes in the nucleus. ΔE int = sum of the average translational energies of all the atoms. K avg = 3 2 nn AkT ΔE int = K avg and from kinetic theory. Therefore, ΔE int = 3 2 nn AkT = 3 2 nrt

26 We know ΔE int We know W We have ΔE int = Q W We now can get the heat Q From the heat we can calculate the specific heat for various processes. For example,

27 Lets calculate the specific heat, C V of an ideal gas at constant volume The internal energy of an ideal monatomic gas like helium and neon is given by the kinetic energy E int = ( 3 2 )nrt and only depends on temperature. We know from experiment in the figure at the right that Q = n C V ΔT, where C V is called the molar specific heat at constant volume The first law of thermo: ΔE int = Q W at constant volume gives ΔE int = Q since W=0

28 Calculate C V But ΔE int = ( 3 2 )nrδt from above Using Q= ΔE int nc V ΔT = 3 2 nrδt C V = 3 R = 12.5J / mol K 2

29 C v for different gases The internal energy for any ideal gas can be written as long as the correct C V is used. E int = nc V T The internal energy and change in internal energy only depends on the temperature change. That is it only depends on the endpoints and not the path. This confirms what we said earlier about the first law. Also note that Q and W depend on the path but the difference Q-W is path independent. See Fig 19-9 in text. 3/2R 12.5 J/mole.K Helium gas 12.5 exp Argon 12.6 exp

30 Specific Heat at Constant Pressure Now C p will be greater because the energy must now do work as well as raise the temperature. Q = nc p ΔT ΔE int = Q W W = pδv pv = nrt ΔE int = nc V ΔT nc V ΔT = nc p ΔT nrδt C V = C p R Another fantastic relation

31 Adiabatic Expansion of an Ideal Gas: Q = 0 Note we do not let any heat get exchanged between system and environment We want to show pv γ = constant γ = C p C V for Q = 0 thermodynamic processes

32 Adiabatic Expansion of an Ideal Gas: Q = 0 ΔE int = Q W Q = 0 de int = W = pdv de int = nc V dt pdv = nc V dt ndt = pdv /C V From the ideal gas law pv = nrt pdv + Vdp = nrdt = n(c P C V )dt pdv + Vdp = (C P C V )ndt = (C P C V ) pdv /C V pdv + Vdp = ( C P C V 1)pdV dp P + C P C V dv V = 0

33 Adiabatic Expansion of an Ideal Gas: Q = 0 dp dv + γ = 0 P V ln P + γ lnv = constant ln PV γ = constant PV γ = constant γ = C p C V

34 PROBLEM 20-58E Show that the speed of sound in a ideal gas for an adiabatic process is v s = Recall v s = B ρ B = V dp dv dp dv = γ p V B = γ p γrt M v s = γ p ρ = γ nrt / V ρ = γ RT M pv γ = constant Also v s = γrt M = γ 3 3RT M = γ 3 v rms γ = C p C V = 5 3 For helium at 300 K, the molecular speed is v rms = 1370 m/s and. Then the predicted speed of sound in helium is 1021 m/s. The measured value is 965m/s

35 Problem 20-61P One mole of an ideal monatomic gas traverses the cycle shown in the figure. (a) Find Q, ΔΕ int, and W for each process. 1 2 constant volume W = 0 ΔE int = Q ΔE int = nc V ΔT = n 3 2 RΔT = (1mol)(1.5)(8.314J / mol K)( )K ΔE int = Q = J

36 Q = 0 ΔE int = nc V ΔT = n 3 2 RΔT = (1mol)(1.5)(8.314J / mol K)( )K ΔE int = J W = Q ΔE int = J

37 3 1 constant pressure C p = C v + R = 3 2 R + R = 5 2 R Q = nc p ΔT = n 5 2 RΔT = (1mol)(2.5)(8.314J / mol K)( )K Q = J ΔE int = nc V ΔT = n 3 2 RΔT = (1.5)(8.314)( ) = J W = Q E int = [( 3.22) ( 1.93)] 10 3 J = J

38 (b) Find the pressure and volume at points 2 and 3. The pressure at point 1 is 1 atm = x 10 5 Pa. At point 1 the volume is determined from V 1 = nrt P 1 = (1)(8.314)(300) = m 3 At point 2, V 2 =V 1 = m 3 p 2 = nrt 2 V 2 = (1)(8.314)(600) = Pa At point 3, P 3 = P 1 = x 10 5 Pa V 3 = nrt 3 P 3 = (1)(8.314)(455) = m 3

39 Equipartition Theroem Equipartition Theorem: Each molecule has 1/2kT of energy associated with each independent degree of freedom All molecules have three ways to move in translation v x,v y,v z Diatomic molecules also have two rotational degrees of freedom as well Polyatomic molecules have three rotational degrees of freedom as well. Monatomic molecules have 0 rotational degrees of freedom How does this effect the specific heats?

40 Specific heats including rotational motion Let f be the number of degrees of freedom E int = f 2 kt = f 2 nrt C V = (3/2) R for monatomic gas (f=3) C V = (5/2) R for diatomic gas (f=5) C V = (3) R for polyatomic gas (f=6) C P = C V + R

41 Below 80 K only translational modes can be excited Above 80 K rotational modes begin to get excited Above 1000 oscillatory modes begin to get excited. At 3200 K the molecule breaks up into two atoms C V for a diatomic gas as a function of T Quantum mechanics is required to explain why the rotational and oscillatory modes are frozen out at lower temperatures. Oscillatory or vibrational modes give 2 more degrees of freedom.

42 What is Entropy S? Two equivalent definitions It is a measure of a system s energy gained or lost as heat per unit temperature. ΔS = S f S i = dq T It is also a measure of the ways the atoms can be arranged to make up the system. It is said to be a measure of the disorder of a system. f i S = k ln W ΔS = k ln W f /W i

43 A system tends to move in the direction where entropy increases For an irreversible process the entropy always increases. ΔS > 0 For a reversible process it can be 0 or increase. ΔS 0

44 Heat Flow Direction Why does heat flow from hot to cold instead of vice versa? Because the entropy increases. What is this property that controls direction? T 2 > T 1 ΔS = S f S i = f i dq T T 2 T 1

45 The statistical nature of entropy Black beans on the right White beans on the left Add some energy by shaking them up and they mix They never will go back together even though energy of conservation is not violated. Again what controls the direction?

46 Assume that entropy is a state property like pressure and volume and only depends on the initial and final state, but not on how it got there. Then the change in entropy is defined as: ΔS = S f S i = f i dq T Note that units are Joule per kelvin and the sign is the same as Q since T > 0 However, the above formula can only be used to calculate the entropy change if the process is reversible.. To find the entropy for an irreversible process and since state functions only depend on the end points, the trick is to replace the irreversible one with a reversible one that has the same end points. Consider the isothermal free expansion of an ideal gas.

47 Calculate the Change in Entropy for an Isothermal Irreversible Free Expansion of an Ideal Gas. No change in temperature.

48 Example:Isothemal Reversible Expansion How do we evaluate S or the change in S? ΔS = S f S i = dq T = 1 T f i f i dq ΔS = Q T To keep the temperature of the gas constant, heat had to be added to the gas while the volume was expanding or else it would have cooled. Since heat was added, Q is +, Hence, the entropy change was + or ΔS > 0 The change in entropy for a free expansion is also given by the above formula since the temperature doesn t change as well.

49 Isothermal expansion of a gas ΔS = Q T ΔE int = nc V ΔT = 0 Isothermal process Q = W From the 1st law ΔS = W T W = f i pdv pv = nrt W = f i nrt dv V W = nrt ln V f V i ΔS = nrln V f V i

50 Example: Four moles of an ideal gas undergo a reversible isothermal expansion from volume V 1 to volume V 2 = 2V 1 at temperature T = 400K. W = V 2 V 2 a) Find the work done by the gas. b) Find the entropy change of the gas? c) Find the entropy change if the process is adiabatic instead of isothermal a) Find the work done. V 2 dv pdv = nrt = nrt ln V 2 V V 1 V 1 W = nrt ln 2V 1 V 1 = nrt ln2 = (4mol)(8.31J / molk)(400k)ln 2 = 9.22x10 3 J

51 b) Find the entropy change of the gas. We just showed that ΔS = W T ΔS = W T = 9.22x103 J 400K = 23.1J / K c) If the expansion is reversible and adiabatic instead of isothermal, find the entropy change of the gas. ΔS = S f S i = f i dq T ΔQ = 0 ΔS = 0

52 For an ideal gas in any reversible process where the temperature and volume may change, the entropy change is given by the following: ΔS = S f S i = f de int = dq dw i dq T dq = de int + dw = nc V dt + pdv dq T = nr dv V + nc dt V T ΔS = nrln V f V i + nc V ln T f T i This holds for all reversible processes for an ideal gas.

53 Summary For isothermal expansion of a gas where the final volume doubles the initial volume, we have the change entropy equal to: ΔS = f i dq T ΔS = nrln V f V i = nrln 2 = k ln2

54 Statistical View of Entropy Boltzman Entropy Equation S = k ln W Now find the change in entropy from The above equation

55 An insulated box containing 6 molecules n 1 =4 n 2 =2 n 1 =3 n 1 =3

56 N=6 Label Percent age n 1 n 2 W S J/K I II III IV V VI VII S = k ln W W = N! n 1!n 2! N!=6x5x4x3x2x1

57 W = N! n 1!n 2! W = 6! 4!2! = 6x5x4x3x2x1 4x3x2x1x2x1 = 6x5 2 = 15

58 This shows the number of microstates available for each configuration Number of Microstates W N=6 molecules Percentage of Molecules in Left Half N=10 22 molecules

59 Statistical View of Entropy Starting from the Boltzman equation S = k ln W show the same result. W i == N! N!0! = 1 N 0 Initial system W f = N! ( N 2 )!( N 2 )! N/2 N/2 Final system ΔS = k lnw f k lnw i 0 N! ΔS = k lnw f = k ln( ( N 2 )!( N ) k ln2 2 )! Same result as that obtained from the gas law.

60 Use Stirling formula ln N! N ln(n) N For very large N N! ΔS = k lnw f = k ln( ( N 2 )!( N ) 2 )! N! ln( ( N 2 )!( N ) = ln(n!) 2 ln(( N 2 )! 2 )!) = ln 2 ΔS = k ln 2 = nrln 2

61 2nd Law of Thermodynamics ΔS 0 If a process occurs in a closed system, the entropy of the system increases for irreversible processes and is constant for reversible processes. Entropy is important in the discussion ofengines and efficiency and Refrigerators,etc. Stop here.

62 Grading for Phys 631 Homework (100 problems) 30% Quizzes (3) 35% Final (1) 29% Spreadsheet 1% Four homework assignments scheduled Fall Homework 1 11:59 PM Sunday August 12 Fall Homework 2 11:59 PM Sunday August 26 Fall Homework 3 11:59 PM Sunday September 9 Fall Homework 4 11:50 PM Sunday September 23 Final Exam 9:00 AM Sept 24 - Sept 26 7:00 PM All Multiple Choice 60 Questions Proctored

63 Classes Fall Registration Aug 15 Phys 605W How things Work I Phys 609W Galileo & Einstein Spring Phys 606W How Things Work II Phys 641W Physics pedagogy Summer 2008 Phys 632 E/M Lecture Phys 635 E/M Lab Phys 633W Modern Physics Online

64 Reminders Check-out on August 2 11:00 AM Turn keys into Conference Services at check-out or you will pay a fine. Return any books to the Library Return Quiz 3 Thank the DEMO techs fo a job well done -Roger and Nicolai

65 The Breaking Broomstick Demo Experiment to demonstrate Inertia First published in 1881 Dramatic-Why does the stick break so violently and leave the glass intact?

66 Breaking Broomstick Demo v LE 2FΔt MASS: m LENGTH: L (1) v CM v CM = FΔt m Resistive torque (2) Iω = L 2 (FΔt) [ τδt] Iω = L 2 mv [ τδt ] CM (3) v LE = v CM L 2 ω

67 Broomstick Breaking Cont. v LE 2FΔt MASS: m LENGTH: v CM Resistive torque LEFT SIDE v LE = v CM L 2 ( L 2 v LE = v CM 3v CM + 6τΔt ml mv CM 1 12 ml ) τδt 2 v LE = 2v CM v LE = 2v CM + 6τΔt ml 1 12 ml 2 RIGHT SIDE v RE = v CM + L 2 ω v RE = v CM + 3v CM v RE = 4v CM v RE = 4v CM 6τΔt ml

68 Apparatus -4 ft long, 7/8in diameter white pine, cedar, or hickory dowel rod or broomstick -Stick pins in each end; cut off heads -Support each pin with a wine glass, coke can, block of wood, etc. Need clearance -Striking stick: Steel 1/2 in diameter and 2ft long -Mark the halfway point of stick so you know where to strike it -Use a hacksaw to etch it around the circumference; avoid stick fracturing due to other weakness. -Raise striking stick and hit the center as hard as you can; follow through